If $\mathbf{c}$ is a simple closed plane curve whose image bounds a region R, and which is traversed counterclockwise, then the area of R is $\int _{c} x dy = -\int _c ydx$, where x and y are the coordinates of the plane. Prove this using Green's Theorem.
Attempt:
Area = $\int\int_R 1dA$.
If we take some P(x,y) and Q(x,y) such that ($\frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y}$) = 1, then $\int\int_R (\frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y})$ dA.
By Green's Theorem $\int\int_R (\frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y})$ dA = $\int_c Pdx + Qdy$.
Now since ($\frac{\partial Q}{\partial x} – \frac{\partial P}{\partial y}$) = 1 there are different possibilities for P(x,y) and Q(x,y), but I don't know how to find all of the different possibilities.
Any help is appreciated.
Best Answer
When you are given problems with two variables but one equation, an easy way to "solve" for one is by eliminating one variable and seeing what you get for the other. In this case, if $Q$ was zero, then you'd have
$$ 1 = -\iint_R \frac{\partial P}{\partial y} dA = \int_C P\,dx $$
and
$$ 1 = -\frac{\partial P}{\partial y}. $$
From this, what is an obvious choice of $P$ based on what you're supposed to end up with? Repeat this for $Q$. Note that I've more or less reverse engineered the problem, but if I just told you what to pick for $P$ and $Q$, you'd be left wondering why other than knowing that it works.